This is a weaker version of a conjecture I asked a few days ago. While that conjecture was shown to be false, upon further work I realized that the answer was not sufficient for my purposes. Here is a reformulated conjecture:
Conjecture: If $a_n$ is a strictly increasing sequence of natural numbers such that $$l_n=\frac{lcm(a_1,a_2,...,a_{n-1})}{a_n}\to 0,$$ then there exists $\lambda>1$ such that $$\lim_{n\to\infty}\frac{\lambda^n}{a_n}=0.$$ The main difference between these two conjectures is that in the previous question, I had set the bound as $\lambda=2$ and attempted to show that the contrapositive (or a stronger version of it) was true.
Again, while the previous question was answered, it provided no insight (at least that I could see) about this question. Should anyone wish to look, the most relevant paper I have found on this matter can be found here.
If
$$l_n=\frac{lcm(a_1,a_2,...,a_{n-1})}{a_n}\to 0,$$
Then
$$\frac{a_{n-1}}{a_n}\to 0,$$
So for every $\epsilon>0$ there exists an $N\in \mathbb{N}$, such that for all $n\geq N$
$a_{n-1}<\epsilon a_n$ which implies $a_{n}>\frac{1}\epsilon a_{n-1}$ $\cdots \cdots(1)$
Now fix any $\lambda>1$.
Now fix $\epsilon$ small enough, so that $\frac{1}\epsilon>\lambda^2$, and for this $\epsilon$ we will get a corresponding natural number $N$ such that $(1)$ holds for all $n\geq N$.
Hence the $(N+k)$th term of the sequence
$$\frac{\lambda^n}{a_n}.$$ is less than $$\frac{\lambda^{N+k}}{\lambda^{2k}\times a_N}.$$, which decreases to $0$ as $k$ increases to infinity as $N$ is fixed.
So I have proved a stronger result which is $$\lim_{n\to\infty}\frac{\lambda^n}{a_n}=0.$$ holds for any $\lambda>1$ if the sequence $l_n$ tends to $0$.